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I have an exercise where I have to calculate and draw a bending moment diagram for a frame structure. The frame structure is similar to the following, image taken from http://structx.com:
How Do I Calculate The Bending Moment Of A Simply Supported Beam
I would like to know, why the bending moment diagrams of vertical beams are made in such a way? Is there an analogy I can use for simple or static graphics?
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The following graph shows a fixed frame subjected to a concentrated load on the horizontal part of the member resulting in a deflected shape.
We draw the tangent lines to the curved shape at the joints B, noting that both curved parts BF and BE are turned away from the respective tangent lines by a rotation angle of $theta = 0$, similar to the cantilever. beam, so we say that these surfaces are under stress. The same thing is observed in the support link A (with respect to the column), and in the loading area.
Now we identify all the points under pressure by realizing that the pressure is caused / rejected by the second, so that we can put the time diagram accordingly – draw the time diagram on the pressure side of the object, which is a general rule. (See chart below)
In the graph above, points E, F, G, and H are inflection points, where, M = 0, and the direction sign changes on either side of the inflection point. (At the bottom of the column, there are several inward forces, as shown in the first picture of your graph, which restrict the outward direction of the legs. Lateral forces produce moments in the opposite direction than moments at the base. . and inward moments of the columns eventually cancel at the cross link.)
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Note that the height-to-width ratio of the frame (h/L) and the column-to-beam moment ratio affect the moment stiffness at the joints, however, for a column with constant geometry throughout. length , o The last (support) time of the column is always half the time at the joint of the column at the end.
I assume you take the “Structural Analysis I” class and ask ahead of time about the topic for the next class. Since it is important to learn the old techniques before trying to predict/measure results with simplifications, I will not go over what I have provided so far, so as not to confuse or contradict your future learning.
Let’s start with the horizontal beam above. if the last link of this metal is in the pin link, you will get the maximum moment in the middle $$M_= frac $$
But because they are fixed after the beam deflects under the load, together the corner rotates slowly and eventually stops balance. Both the frame and column are slightly bent by rotating the joint. In this position, the moment at the end of the beam and at the top of the column is equal to the moment at the center of the beam is reduced.
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Now part of this moment will be transferred to the base plate (pull element), due to the elastic property of the column.
There are many ways to calculate how time is redistributed, but the Hardy Cross method is the simplest way to calculate subframes. [Source Wiki
By clicking “Accept all cookies”, you agree that Stack Exchange may store cookies on your device and provide information in accordance with our Cookie Policy. in Figure 7, draw a shear force diagram (SFD) and a bending moment diagram (BMD). 15 kN A 3 m 2 m Figure 7 8
QUESTION 4 a) Calculate the shear and bending strength of steel under concentrated load as shown in figure 7, and draw the shear force diagram (SFD) and bending moment diagram (BMD). 15 kN A 3 m 2 m Figure 7 8
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